Question

makenzie throws a ball from the top floor of the physics building. the function $h(t)=-16(t - \frac{1}{2})^2+100$ models the height, $h$, of the ball above the ground in feet as a function of time $t$ measured in seconds since the ball was released.
a) when will the ball hit the ground?
the ball will hit the ground after
seconds.
b) how high is the ball from the ground when it is released?
the ball was
feet above the ground before its initial released time.
c) how long will the ball have traveled in the air before it stops rising and starts falling?
the ball was in the air for
seconds.
d) what will be the maximum height of the ball?
the ball will have reached a maximum height of
feet above the ground.

Explanation:

Step1: Find when the ball hits the ground (A)

Set $h(t)=0$. So, $0=-16(t - \frac{1}{2})^2+100$.
First, rearrange the equation:
$16(t - \frac{1}{2})^2=100$.
Then, $(t - \frac{1}{2})^2=\frac{100}{16}=\frac{25}{4}$.
Take the square - root of both sides: $t-\frac{1}{2}=\pm\frac{5}{2}$.
Case 1: $t-\frac{1}{2}=\frac{5}{2}$, then $t=\frac{5 + 1}{2}=3$.
Case 2: $t-\frac{1}{2}=-\frac{5}{2}$, then $t=\frac{-5 + 1}{2}=-2$. Since time $t\geq0$, we discard the negative solution. So, $t = 3$ seconds.

Step2: Find the initial height (B)

The ball is released at $t = 0$. Substitute $t = 0$ into $h(t)$:
$h(0)=-16(0-\frac{1}{2})^2 + 100=-16\times\frac{1}{4}+100=-4 + 100=96$ feet.

Step3: Find when the ball stops rising (C)

The function $h(t)=-16(t - \frac{1}{2})^2+100$ is in vertex - form $y=a(x - h)^2+k$, where $(h,k)$ is the vertex of the parabola. For a parabola $y = a(x - h)^2+k$ with $a\lt0$ (here $a=-16\lt0$), the vertex represents the maximum point. The $x$ - coordinate of the vertex of the parabola $y=a(x - h)^2+k$ is $x = h$. In our function $h(t)=-16(t - \frac{1}{2})^2+100$, the value of $t$ at the vertex is $t=\frac{1}{2}$ seconds.

Step4: Find the maximum height (D)

The maximum height occurs at the vertex of the parabola. The function $h(t)=-16(t - \frac{1}{2})^2+100$ has a vertex at $(t,h)=( \frac{1}{2},100)$. So the maximum height is $h = 100$ feet.

Answer:

A. 3
B. 96
C. $\frac{1}{2}$
D. 100